Question

I am a three digit odd number sum of my digits is a prime number iam divisible by only two prime number my tens digit is greater than ones place my ones digit is tour less than hundreds digit my tens digit is 1 more than the difference of other two digits who am i

Answer 1

Answer:

221

Step-by-step explanation:

2+2+1=5

221 is divisible by 13 and 17

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Answer 2

Correct Question :- I am a three digit odd number. sum of my digits is a prime number . i am divisible by only two prime number. my tens digit is greater than ones place by ones digit . my tens digit is 1 more than the difference of other two digits . who am i ?

Solution :-

we have given that :-

• a three digit odd number.
• sum of digits is a prime number .
• only divisible by two prime number.
• tens digit is greater than ones place by ones digit .
• tens digit is 1 more than the difference of other two digits .

Case 1 :- three digit number is odd.

So, unit digit will be either 1,3,5,7 or 9.

Case 2 :- tens digit is greater than ones place by ones digit .

So, tens digit can be 2, 4, 6 and 8.

conclusion from both case :- unit digit cant be 9. ( than tens digit = 9 + 1 = 10 ≠ Not possible.)

case 3 :- tens digit is 1 more than the difference of other two digits .

Lets check all Possible tens digits now :-

1) if tens digit is 8 , than unit digit will be 7 .

Than,

Let hundreds digit is x .

So,

→ tens digit - 1 = x - 7 or 7 - x

→ 8 - 1 = x - 7 or 7 - x

→ 7 = x - 7 or 7 - x

→ x = (7 + 7) or (7 - 7)

→ x = 14 or 0 .

Both Not possible.

2) if tens digit is 6 , than unit digit will be 5 .

Than,

Let hundreds digit is x .

So,

→ tens digit - 1 = x - 5 or 5 - x

→ 6 - 1 = x - 5 or 5 - x

→ 5 = x - 5 or 5 - x

→ x = (5 + 5) or (5 - 5)

→ x = 10 or 0 .

Both Not possible.

3) if tens digit is 4 , than unit digit will be 3 .

Than,

Let hundreds digit is x .

So,

→ tens digit - 1 = x - 3 or 3 - x

→ 4 - 1 = x - 3 or 3 - x

→ 3 = x - 3 or 3 - x

→ x = (3 + 3) or (3 - 3)

→ x = 6 or 0 .

6 can be Possible.

So,

→ Three digit number will be :- 643 .

Now,

→ sum of digits = 6 + 4 + 3 = 13 = Prime number .

But also,

→ 643 = is a prime number and it is not Multiple of two prime numbers .

Hence,

Not Possible.

4) Now tens digit can be 2 , than unit digit will be 1.

Let hundreds digit is x .

So,

→ tens digit - 1 = x - 1 or 1 - x

→ 2 - 1 = x - 1 or 1 - x

→ 1 = x - 1 or 1 - x

→ x = (1 + 1) or (1 - 1)

→ x = 2 or 0.

So,

→ Three digit number will be :- 221.

Now,

→ sum of digits = 2 + 2 + 1 = 5 = Prime number .

and also,

→ 221 = 13 * 17 = Multiple of two prime numbers 13 and 17.

Hence, The required Three digit number is 221.